Question

the value of k for which the line
represents are coincident is 2x+3y+7=0
8x +12y+k=0​

Answer 1

Answer:

i think 28 is your right answer

Answer 2

SOLUTION

TO DETERMINE

The value of k for which the line represents are coincident is 2x + 3y + 7 = 0 and 8x + 12y + k = 0

CONCEPT TO BE IMPLEMENTED

For the given two linear equations

$\displaystyle \sf{ a_1x+b_1y+c_1=0 \: and \: \: a_2x+b_2y+c_2=0}$

Consistent :

One of the Below two condition is satisfied

1. Unique solution :

$\displaystyle \sf{ \: \frac{a_1}{a_2} \ne \frac{b_1}{b_2} }$

2. Infinite number of solutions ( Coincident line )

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \: \frac{c_1}{c_2}}$

Inconsistent :

NO solution

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \: \frac{c_1}{c_2}}$

EVALUATION

Here the given system of equations are

2x + 3y + 7 = 0 and 8x + 12y + k = 0

Comparing with the equation

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 we get

a₁ = 2 , b₁ = 3 , c₁ = 7 and a₂ = 8 , b₂ = 12 , c₂ = k

Now the given system of equations is coincident

Then we have

$\displaystyle \sf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \: \frac{c_1}{c_2}}$

$\displaystyle \sf{ \implies\: \frac{2}{8} = \frac{3}{12} = \: \frac{7}{k}}$

$\displaystyle \sf{ \implies\: \frac{1}{4} = \frac{7}{k}}$

$\displaystyle \sf{ \implies\: k = 4 \times 7}$

$\displaystyle \sf{ \implies\: k = 28}$

FINAL ANSWER

Hence the required value of k = 28

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